\[ \begin{align}
x+\log{(x)} = y-1\\
y+\log{(y-1)} = z-1 \\
z+\log{(z-2)} = x+2 \\
\end{align}\]
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There is only one solution set \(\left(x,y,z\right)\) to the system of equations above. Evaluate \(x+y+z\).

Note: Take the base of the logarithms as 10.

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